Problem: The sides of an isosceles triangle are $\cos x,$ $\cos x,$ and $\cos 7x,$ and its vertex angle is $2x.$  (All angle measurements are in degrees.)  Enter all possible values of $x,$ separated by commas.
Explanation: Note that angle $x$ must be acute.

If we drop an altitude from the vertex of the isosceles triangle, then we obtain two right triangles, where one of the angles is $x,$ the opposite side is $\frac{\cos 7x}{2},$ and the hypotenuse is $\cos x.$  Hence,
\[\sin x = \frac{\frac{\cos 7x}{2}}{\cos x} = \frac{\cos 7x}{2 \cos x}.\]Then $\cos 7x = 2 \sin x \cos x = \sin 2x.$  We can write this as $\cos 7x = \cos (90^\circ - 2x).$  Then the angles $7x$ and $90^\circ - 2x$ must either add up to a multiple of $180^\circ,$ or differ by a multiple of $90^\circ.$

In the first case,
\[7x + 90^\circ - 2x = 180^\circ k\]for some integer $k.$  Then
\[x = 36^\circ k - 18^\circ.\]The only acute angles of this form are $18^\circ$ and $54^\circ.$  Furthermore, if $x = 18^\circ,$ then $\cos 7x = \cos 126^\circ < 0.$  We check that $x = 54^\circ$ works.

In the second case,
\[7x - (90^\circ - 2x) = 180^\circ k\]for some integer $k.$  Then
\[x = 20^\circ k + 10^\circ.\]The only acute angles of this form are $10^\circ,$ $30^\circ,$ $50^\circ,$ and $70^\circ.$  Again, $\cos 7x < 0$ for $x = 30^\circ$ and $70^\circ.$  We check that $10^\circ$ and $50^\circ$ work.

Thus, the possible values of $x$ are $\boxed{10^\circ, 50^\circ, 54^\circ}.$